JEE Main 2020 (Online) 3rd September Evening ShiftQuestion 1
Let $e_1$ and $e_2$ be the eccentricities of the ellipse, $\frac{x^2}{25}+\frac{y^2}{b^2}=1$ ($b<5$) and the hyperbola, $\frac{x^2}{16}-\frac{y^2}{b^2}=1$ respectively satisfying $e_1e_2=1$. If $\alpha$ and $\beta$ are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair $(\alpha,\beta)$ is equal to :
Hyperbola
Solution
$e_1^2=1-\frac{b^2}{25}$, $e_2^2=1+\frac{b^2}{16}$; $e_1e_2=1$ gives $\left(1-\frac{b^2}{25}\right)\left(1+\frac{b^2}{16}\right)=1\Rightarrow b^2=9$. Ellipse: $c=4$, $\alpha=8$; hyperbola: $c=\sqrt{16+9}=5$, $\beta=10$. So $(\alpha,\beta)=(8,10)$.